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I was playing around with modular arithmetic when I noticed something.

\begin{align} g^k &= a \bmod n\\ g^{k+0.5(n-1)} &= b \bmod n \end{align}

Then $a + b = n$, so you can also write

$$g^{k+0.5(n-1)} = n - a \bmod n$$

This seems to hold if $n$ is prime and $g$ is the primitive element of $n$, according to randomly trying sets of $n$ and $g$. I have tried to see the theory behind this, but was not able to do so. None of the properties (Wikipedia: Modular arithmetic properties) of modular arithmatic seems to relate to this. Can someone explain why this holds?

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If you know the Fermat's little theorem, that is, $g^{n-1} = 1 \mod n$, then you can understand what you found.

The point is that

$$b^2 = g^{2k + n-1} = g^{2k}g^{n-1} = g^{2k} = a^2 \mod n$$

Therefore, $b = a$ or $b = -a$ in $\mathbb{Z}_n$. But $b = -a$ means that $b = n-a$, since $n = 0$ modulo $n$.

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  • $\begingroup$ Thank you for explaining. This shows that the first 0.5n numbers of the order influences the second 0.5n numbers, right? So, there is some structure in the 'random' modular arithmetic $\endgroup$ – user71649 Sep 4 at 15:26
  • $\begingroup$ I don't understand what you mean by the "first 0.5n numbers of the order", sorry. $\endgroup$ – Hilder Vítor Lima Pereira Sep 5 at 6:40

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